Answer:
Part A: f(x) = 4·x² - 7·x-15 = (x - 3)·(4·x + 5)
Part B: The x-intercepts are x = 3 and -1.25 #
Part C: As x approaches ∞, y approaches ∞, as x approaches -∞, y approaches ∞
Part D: The graph is a u-shaped quadratic equation graph passing though the x-intercept points -1.25 and 3 on the x-axis and the y-intercept point -15 on the y-axis. Both sides of the symmetrical graph curve extending to infinity
Step-by-step explanation:
The function 4·x² - 7·x-15
To factorize the expression, we have at the 0 factors;
![x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}](https://tex.z-dn.net/?f=x%20%3D%20%5Cdfrac%7B-b%5Cpm%20%5Csqrt%7Bb%5E%7B2%7D-4%5Ccdot%20a%5Ccdot%20c%7D%7D%7B2%5Ccdot%20a%7D)
Where;
a = 4, b = -7, c -15, substituting gives;
![x = \dfrac{-(-7)\pm \sqrt{(-7)^{2}-4\times 4\times (-15)}}{2\times 4} = \dfrac{7\pm 17}{8} = 3 \ or -1.25](https://tex.z-dn.net/?f=x%20%3D%20%5Cdfrac%7B-%28-7%29%5Cpm%20%5Csqrt%7B%28-7%29%5E%7B2%7D-4%5Ctimes%204%5Ctimes%20%28-15%29%7D%7D%7B2%5Ctimes%204%7D%20%3D%20%20%5Cdfrac%7B7%5Cpm%2017%7D%7B8%7D%20%20%3D%203%20%5C%20or%20-1.25)
Which gives;
(x - 3)·(x + 1.25) = (x - 3)·(4·x + 5)
f(x) = 4·x² - 7·x-15 = (x - 3)·(4·x + 5)
Part B: The x-intercepts are the point where y = 0, which are x = 3 and -1.25 as shown above
Part C: As x approaches ∞, y approaches ∞, as x approaches -∞, y approaches ∞
Part D: With the the nature of the function as y approaches ∞ and -∞ the graph is a u-shaped graph of a quadratic equation passing though the points -1.25 and 3 on the x-axis and the point -15 on the y-axis.